Problem: Let be cocompact lattice in locally compact. Given , what can we say about (up to compact groups) ?

Mosher-Sageev-White: If is free, must act on a tree.

Bader-Furman-Sauer: If is a lattice in a group in a wide class, must be .

I focus on abelian by cyclic groups. Such groups admit presentations of the form for some integer matrix with non vanishing determnant.

Theorem 1 1. If det(M)=1 and no unit eigenvalue,

or

2. all eigenvalues are in absolute value,

then is a subgroup of for some rational , where is the absolute Jordan form of .

Example 1. Then is a lattice in .

Example 2. Then .

Outline : next,

– I describe and its isometry group (focussing on case 2).

– I outline the proof.

– I mention case 1 and case where has eigenvalues and .

1.

It is made of two pieces, a tree and a negatively curved Lie group. Let . Let be the regular tree with degree , oriented so that there are 1 incoming and outgoing edges at each vertex. This defines a height function . Let be the semi-direct product where acts on by the matrices . This comes with a height function as well. Let

with the path metric.

Example 3 : For , is homeomorphic to , in which factors are horocycles in hyperbolic planes.

is a hyperbolic metric space, whose ideal boundary is the one point compactification of where is the field of -adic numbers, and has the Rickmann rug metric (product of snowflaked lines). Every isometry of acts on boundary by a pair of similarities , a similarity of , a similarity of , with reciprocal scaling factors.

2. Proof sketch

If is cocompact in , then becomes (up to compact groups) a group of uniform quasiisometries of .

Theorem 2 (Farb-Mosher)

Theorem 3 (Mosher-Sageev-Whyte) Every uniform subgroup of can be conjugated in for some depending on .

Theorem 4 Every uniform subgroup of can be almost conjugated in . Furthermore, if is scalar, I can remove the almost.

Thus can be conjugated into . Using uniformity, one then clears up the scaling factors and conclude that is contained in Isom for some rational .

Note if is not diagonal, a uniform biLipschitz cyclic group of is has a triangular form involving non diagonal translations ; I call this almost-similarity.

3. Case 1:

If , is a solvable Lie group. Its quasi-isometry group is understood again (Eskin-Fisher-Whyte).

However, the analysis of biLipschitz maps in the non diagonalisable case does not work any more.